U.U.D.M. Project Report 2020:42

Examensarbete i matematik, 15 hpHandledare: Ingemar KajExaminator: Martin HerschendSeptember 2020

Department of MathematicsUppsala University

Exploring strategies in Monopoly usingMarkov chains and simulation

Albert Nilsson

Contents

1 Introduction 11.1 Introduction to the game . . . . . . . . . . . . . . . . . . . . . 11.2 Delimitations . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Purpose and problem formulation . . . . . . . . . . . . . . . . 4

2 Theory 42.1 Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1.1 Stationary distribution for discrete Markov chains . . . 5

3 Modelling monopoly as a Markov chain 73.1 State space and description of model . . . . . . . . . . . . . . 73.2 Transition probabilities . . . . . . . . . . . . . . . . . . . . . . 8

3.2.1 Dice . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2.2 Cards . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2.3 Transition matrix . . . . . . . . . . . . . . . . . . . . . 11

3.3 Stationary distribution and result . . . . . . . . . . . . . . . . 12

4 Extended model of Monopoly 184.1 The structure of the simulation . . . . . . . . . . . . . . . . . 184.2 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Summary and discussion 20

6 Further exploration 22

1 Introduction

Monopoly is a classic that almost everyone has played or at least heard of.It is a game which bring out joy and disputes in as well as family as friends.There seems to be a universal idea that the more expensive the better andthat buying much will generate great result and it is easy to see why. If youmake a big investment it is intuitive to expect to be rewarded. However isthis in fact the case? In this project Monopoly was examined and an attemptto find the best possible strategy was made. It was investigated if all streetsare equally probable, how the expected return change when buying houses,which colour will give you the most bang for the buck and which are the bestcolours to obtain.

The first part of this paper the focus lie in finding the probabilities oflanding at each tile. That is, one player will be imagined walking around theboard without making any decision or transactions. The second part will bemore of an experimental part as to the first being more theoretical. This partincludes simulating players buying different colours and see which works thebest. One could say that this kind of test the conclusions of the first part.

1.1 Introduction to the game

Monopoly is a game about buying streets, collecting money, ruining youropponents and is won by being the last to have money left. The board gameconsist of forty tiles and each player traverse these by rolling two dice andmove accordingly. If a player rolls a double, namely rolls two of the samenumber, she plays her turn as usual but after making decisions and payingeventual fees she plays another turn. If a player rolls three doubles in a rowthe player goes to jail. While in jail you have three options to get out, eitheryou use a get-out-of jail card, pay 1000 kr or roll a double. Only one optioncan be chosen each turn meaning that you cannot try rolling the dice andthen choose another option in the same turn. A player can spend a maximumof 3 turns in jail before being released. If a player has been in jail 3 roundsshe is transferred to visit-jail and can on next turn play as usual.

• In preparation to start one player is chosen as a banker whose purposeis the distribution of assets between the bank and the players. Eachplayer start at the tile GO with 30 000 kr.

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• On a players turn the player must roll her dice. If the player wishes totrade, improve buildings or alike it must be done before she roll.

• If a player lands on a property not yet owned, the player is asked tobuy the property. If she accept the player pays the bank and obtainsthe property. Otherwise the property goes to auction where the highestbidding player obtains the property and pays the given amount to thebank.

• If a player lands on a street that her self own nothing happens. Howeverif landing on another players street the player must pay rent if theowner notice. If however one lands on someones utility the player mustpay 80 times what they rolled. If all utilities are owned by same ownerit is instead 200 times.

• If a player lands on a community chest or chance tile the player mustdraw a card, do what the cards says and then put the card at thebottom of the pile.

• If a player lands on the tile luxury tax they must pay 2000 kr if insteadlanding on income tax the player must pay 4000 kr.

• Landing on the go-to-jail tile means that the player transfers directlyto jail without passing GO. However landing on the jail tile means justvisiting and no penalties apply.

• If passing GO the player is rewarded with 4000 kr from the bank.

• If a player owns every street in one colour they may buy up to 4 housesor 1 hotel (hotels can only be bought after owning 4 houses on thatstreet) for each property belonging to that colour. It does not haveto be the players turn nor does the player have to be on the propertythat she choose to place a house/hotel on. However the house/hotelsmust be distributed evenly in a colour meaning that a house can onlybe placed on the street with fewest amount of house on. A house canbe sold back to the bank for half of its initial cost.

• A property may be mortgaged for half of its price and cannot be devel-oped nor can it collect rent while in this state. To end the mortgaged

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the player must pay the mortgage price with 10% interest. If a mort-gaged property is traded between players the new player must pay 10% interest directly and then the same rules applies as above.

1.2 Delimitations

To make Monopoly more comprehensible to model some limitations of thegame rules were made:

• Mortgaging properties or selling houses/properties is not allowed.

• Only 2-player games will be considered.

• Get-out-of jail cards and paying to get out will not be considered.

• Utilities will charge 7 times 80/200 instead of the sum of the dice times80/200

• Trade will not be considered.

• No auction of the property if a player decide not to buy.

• The cards are put back randomly.

The limitations is purely motivated by that they reduced the complexity ofthe game and made it easier to simulate. For utilities instead of charging thedice roll times 80/200 the sum of the dice roll were replaced with 7 since itis the mean of the sum of 2 dice. Trading was not considered since in thesecond part of the paper where it was relevant, only 2-player games wereconsidered and then trade does not make sense. If a player want to makean efficient trade then in some sense, she wants to reduce the probability ofgetting ruined. But intuitively in doing so the player have to increase anotherplayers probability of losing. Then of course if there are only two players,someone will make a bad trade and should not trade. If a player does not buya property it will not be auctioned out. The reason for this is that it simplydoes not make sense to hold an auction for 1 person. Lastly the reason forputting the cards back randomly instead of at the bottom of the pile is tosimplify the distribution of the cards and also to obtain time-homogeneity ofthe Markov chain.

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1.3 Purpose and problem formulation

The main purpose of this essay is to try to investigate if there exist a wayto play Monopoly good or if it is simply a game of chance. If it is morethan a game of luck then how should a player behave? This leads to themain question of this essay: Does an optimal strategy of Monopolyexist? However since this is a difficult and rather broad question it will bedivided in to sub-questions. It should be noted that these questions does notnecessary capture the larger question as a whole but is important for drawingconclusion towards the goal.

• Do the probabilities of the tiles differ and which colour is most probableto land on?

• What is the optimal amount of houses to own for each street?

• Which colour generates most money?

• Which are the best colours to obtain?

2 Theory

The main purpose of this chapter is to build up basic theory about Markovchains and stating requirements for existence of stationary distribution fordiscrete Markov chains. The reason for this is that as will be seen later itis possible to model Monopoly using a Markov chain with discrete time anddiscrete state space. By finding the stationary distribution it is possible tofind the probability of landing at each tile independent of where you currentlystand.

2.1 Markov Chains

In order to define a Markov chain the concept of stochastic processes mustfirst be introduced.

Definition 2.1. Given a probability space (Ω,F , P ), a stochastic process isa function X : [0,∞]× Ω −→ R. We denote the function as Xt = X(t, ω) [1].

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Definition 2.2. A Markov Chain is a stochastic process satisfying

P (Xn = j|Xn−1 = in−1, ..., X0 = i0) = P (Xn = j|Xn−1 = in−1)

This is called the Markov property [2].

This paper will only consider Markov chains with discrete state spaceand discrete time. Furthermore the focus will lie solely in time-homogeneousMarkov chains. This means that the probability of going from state i to jdoes not depend on n, that is

P (Xn = j|Xn−1 = in−1) = P (X1 = j|X0 = in−1)

.Note that

pij = P (Xn = j|Xn−1 = i)

is called the transition probability from state i to j. Now suppose Xtt∈Tis a Markov chain. Then matrix P where the entry (i,j) equals pij such thati, j ∈ S is called the transition matrix of the Markov chain. This matrix hasthe property that

∑j∈S

pij = 1 ∀i ∈ S

and pij ≥ 0 ∀i, j ∈ Smaking P a stochastic matrix [2].

2.1.1 Stationary distribution for discrete Markov chains

The stationary distribution can be intuitively thought of as the distributionin the long run that is, the probabilities of reaching the different states in-dependent of where you start. More formally, a stationary distribution π isthe distribution satisfying the equation

π = πP

[2]. To reach the theorem that tells us when a stationary distribution existit is necessary to introduce some properties and classifications of Markovchains.

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Definition 2.3. A Markov chain is said to be irreducible if pij > 0 for alli, j in the chain.

Definition 2.4. Let Tn be the first time of return to state n. Then the staten is called recurrent if

P (Tn <∞) = 1

Moreover it is called positive recurrent if

E(Tn) <∞

[2] This means that a recurrent state will with probability 1 be revisitedafter a finite number of step. Now it is possible to state the following usefultheorem:

Theorem 2.1. Suppose Xn is an irreducible Markov chain then

1. If one state is positive recurrent then all states are positive recurrent.

2. All states are positive recurrent ⇐⇒ A stationary distribution exist.

Apart from this we also have the following useful theorem.

Theorem 2.2. Suppose Xn is a Markov chain defined on a finite statespace S. Then at least one j ∈ S is recurrent and every recurrent state ispositive recurrent. [3]

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3 Modelling monopoly as a Markov chain

3.1 State space and description of model

To get a better understanding of the game a reasonable first step of analysiswas to solely look at the board and finding the probabilities of landing oneach tile. Therefore the first model of the game excluded money as well asany decisions and instead focused on capturing how one player move aroundon the board. To motivate why this might be interesting, a player wins thegame by ruin its opponents. This is essentially done by establishing streetswhere the player can charge the opponents for rent. But in order to claimrent a player must stand on your tile and the players will of course tend tostand more often on the most probable tiles.

To get a grip of how the probabilities of the tiles arise it is key to realizehow a player can move on the board. There are two primary ways, rollingthe two dice and drawing cards. An additional feature of the dice is rollingdoubles. Now in order to model this as a Markov chain its tempting to justsay that there are 40 states in the state space, one for each tile. But by givingit a quick ponder one realize fast that it is not sufficient. The reason for this isthat this fails to incorporate how many doubles a player has thrown and thusfails to satisfy the Markov property. For example, a player who has throwntwo doubles has in its state remarkably higher probability to go to jail thana player that has not yet rolled a double standing in the same position. It issimilar when a player is in jail. Then instead a player who has stood thereone turn can only get out by rolling a double while a player who has stoodthere for three turns will get out with a probability of 1 and is in the samestate as a player just visiting jail. With that said to get a correct model thestate space must keep track of how many doubles the player has thrown foreach non-jail tile and how many doubles the player has not thrown if in jail.This gives that the state space will have 3 · 39 + 3 = 120 states. Note thata player has 0 probability of ending a turn on the go-to-jail tile since theplayer will be directly sent to jail. However that position (30) will be used tokeep track of people in jail and the jail tile will only keep track of visitings.Each state will be represented as (doubles, position, timeJail). The variableposition will be between 0 and 39 and doubles and timeJail will be between0 and 3. Doubles and timeJail can not be non-zero at the same time since ifa player is in jail and throw a double the player will be released and timeJailwill be set to 0. Now it is possible to presented all the states as following:

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doubles position timeJail0 k 01 k 02 k 00 30 00 30 10 30 2

Table 1: All possible states. k ∈ 0, ...39 \ 30

3.2 Transition probabilities

Now that it is clear which states the model has it remains to find the tran-sition probabilities between these states. It should be noted that there are120 · 120 = 14400 possible transitions so Matlab was used to create the ma-trix and find the stationary distribution. Since movement depends on thetwo dice and the cards, the probability distribution of these must be found.

3.2.1 Dice

Each time it is a players turn she throws two dice. The sum of these arethe amount of steps the player walk. The sum can vary between 2 and12 and the probability can easily be calculated. Although since a playerstransitions depends on if she rolls a double or not its important to separatethem. Throwing a specific double is always done with a probability of 1/36since there are exactly one unique pair where the eyes of the dice are thesame for every number on a die. Furthermore the sum of a double can onlybe even. The distribution looks as following:

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Sum Probability (non-double) Probability (double)2 0 1/363 1/18 04 1/18 1/365 1/9 06 1/9 1/367 1/6 08 1/9 1/369 1/9 010 1/18 1/3611 1/18 012 0 1/36

Table 2: Probability distribution of the sum of two dice.

From each non-jail state there are three transitions that can happen;A player does not roll a double and transition to the new position with 0doubles, a player rolls a double and goes to the new position with one moredouble or a player rolls its third double in a row and goes to jail. If insteadin jail a player can either roll a double and get out, stay another turn or getout because she has been there for 3 rounds. Note that the matrix becomes120× 120 since there are 120 states.

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Current state New State Transition prob(d,pos,0) (0,pos+3 mod 40,0) 1/18(d,pos,0) (0,pos+4 mod 40,0) 1/18(d,pos,0) (0,pos+5 mod 40,0) 1/9(d,pos,0) (0,pos+6 mod 40,0) 1/9(d,pos,0) (0,pos+7 mod 40,0) 1/6(d,pos,0) (0,pos+8 mod 40,0) 1/9(d,pos,0) (0,pos+9 mod 40,0) 1/9(d,pos,0) (0,pos+10 mod 40,0) 1/18(d,pos,0) (0,pos+11 mod 40,0) 1/18(d,pos,0) (d+1,pos+doubleValue mod 40,0) 1/36(2,pos,0) (0,30,0) 1/6(0,30,t) (0,10+doubleValue,0) 1/36(0,30,t) (0,30,t+1) 5/6(0,30,3) (0,10,0) 1

Table 3: Possible transitions by throwing two dice. Note that in this tabled ∈ 0, 1, 2, t ∈ 0, 1 and doubleV alue ∈ 2, 4, 6, 8, 10, 12

3.2.2 Cards

There are two types of cards, community chest cards and chance cards. Mostof the cards in both involve some sort of money transaction but since moneyis excluded from the model these will be transition back to the same state.

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Current state New State Transition prob(d,chancePos,0) (d,chancePos,0) 1/2(d,chancePos,0) (d,11,0) 1/16(d,chancePos,0) (d,0,0) 1/16(d,chancePos,0) (d,24,0) 1/16(d,chancePos,0) (d,1,0) 1/16(d,chancePos,0) (0,30,0) 1/16(d,chancePos,0) (d,15,0) 1/16(d,chancePos,0) (d,39,0) 1/16(d,chancePos,0) (d,chancePos-3 ,0) 1/16(d,chestPos,0) (d,chestPos,0) 14/16(d,chestPos,0) (d,0,0) 1/16(d,chestPos,0) (d,30,0) 1/16

(d,pos,j) (d,pos,k) 1

Table 4: Transitions by drawing cards. chancePos ∈ 7, 22, 36,chestPos ∈ 2, 17, 36 and pos ∈ 1, ..., 39 \ cardPos, where cardPos =2, 7, 17, 22, 33, 36

Note that the last entry of the table is the probability of staying in thesame position when not standing on a card tile. This is clearly 1 since youcannot draw a card there. Moreover the transition matrix here will be of size120 × 120 with ones on the diagonal for the non-card tiles and for the cardtiles it follows the distribution shown in table 4.

3.2.3 Transition matrix

In order to obtain the transition matrix for the whole model the transitionmatrix of the cards must be combined with the transition matrix of the twodice. To do this its important to realize that a player first throws the dicethen draw a card and that what card a player draws is independent of whatthe sum of the dice was. This means that the probability of going from statei to j will be the transition probability of rolling the dice times the transitionprobability of the cards. Moreover if not standing on a card tile the transitionprobability will follow the dice distribution. Since in this essay probabilityvectors will be row vectors the transition matrix becomes P = DC where Dis the transition matrix for the dice and C for the cards. Note that we infact can multiply the matrices together since both are of size 120× 120.

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3.3 Stationary distribution and result

The chain is irreducible. This can be seen by noting that given every positionit is possible to walk around the board until every other position is reached.Since it is irreducible and finite we have by the first part of Theorem 2.1together Theorem 2.2 that the chain only has positive recurrent statesand therefore by part 2 of Theorem 2.1 there exist a unique stationarydistribution.

To find the stationary distribution the equation π = πP has to be solvedfor a π with only positive components that sum to one. Since π = πP is theequivalent to that π is the eigenvector of P corresponding to the eigenvalue1 it suffices to find that eigenvector.

Figure 1: Probability of landing on each tile.

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It should be noted that Chance 1 refers to the chance tile on the firstrow after GO and so on. Furthermore looking at the probabilities one canquickly say some things about the game. The first most obvious is the factthat all streets are not equally probable. This is key since this reinforces thesuspicion that Monopoly is a game where it is possible to find ways to makegood decisions. However it should be noted that the differences is small.Hamngatan and Centrum which have the highest and lowest probability ofthe buyable tiles only differs with 0.95 %. This means that for a game with200 rounds you could expect about 2 more visits to Hamngatan than Cen-trum. Another fact that catch the eye is the high probability of the going tojail. This make sense since the game rules basically revolves around going tojail. If a player rolls 3 doubles in a row, draws a go-to-jail card or land on thego-to-jail tile she goes to jail. It is also notable that all of the orange streetshave high probability and could partly be a consequence of being able totransition to them directly from jail. However since you cannot buy housesuntil you own all streets of the same colour its interesting to consider theprobability of the colours as a whole.

Figure 2: Total probability of landing on each colour

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As seen in the picture orange and red are the most probable and brownand dark blue are the least probable. It should be noted that brown and darkblue are the only colours with 2 streets but they also contain some of thestreets with lowest probabilities with Centrum (Dark blue) being the buyable-tile with the lowest probability. Now unfortunately it is not enough to justlook at the probabilities. One also have to consider the potential return ofinvesting in different colours. A player surely want the most bang for thebuck. This leads to consideration of houses. How does the expected returnchange when a player invest in houses? Return here refers to the expectedamount of money it will generate on a turn divided by cost. Immediately by

Figure 3: How the expected return change when developing houses. For 0houses it is only the expected return.

looking at figure 3 one can notice some interesting things. Firstly there seemsto be a peak in the change of expected return at 3 houses. This means thatgoing from 2 to 3 houses has the largest increase of expected return. Orangehas especially large expected increase with as much as 1.7 %. Otherwise

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it seems to be the fact that the expected return increase as you buy morehouses. There are however two expectations, dark blue and green. Whenincreasing from 3 and 4 houses per street a player actually get less for whatshe pays for. In addition to see how the return changes it is also interestingto consider how many opponent turns until profit is expected to be made.

colour no house 1 house 2 houses 3 houses 4 houses 5 housesbrown 469 341 165 73 51 41blue 370 218 103 43 35 30pink 295 198 93 41 37 35

orange 237 146 70 32 29 27red 215 147 73 35 35 34

yellow 240 150 68 36 36 36green 251 162 76 43 43 44

dark blue 197 139 64 35 36 36

Table 5: Number of opponent turns to profit.

This corresponds as expected well with figure 3. For example for greenthe expected turns to profit increase when going from 4 houses to 5 whichis reasonable since we saw in figure 3 that the expected return actually de-creased. For the other colours the amount of turns either remained the sameor decrease. The reason why dark blue does not have an increase in turnsis due to that the decrease in expected return is so small. Green also hasthe most amount of turns to profit for 5 houses. Orange has the lowest ex-pected rounds to profit for 3 houses or more. But apart from when a colourmakes profit it is also of interest to considered how much profit each colouris expected to generate over time. Time in this case simply means how manyturns the opponents have. The expected return will be restricted to 5 housessince it was the investment with greatest expected return for the majority ofthe colours.

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Figure 4: How the expected profit evolves over opponents turns.

From figure 4 we can see that orange is one of the streets that is expectedto generate most profit with 5 houses. If comparing orange to dark blueit can be seen that orange is cheaper to invest in and have more expectedprofit and thus making orange strictly better than dark blue with 5 houses.Red and yellow seems to be about the same and green generates slightlybetter profit however it cost 17 400 kr more than the second most expensive(yellow) and it is a small difference between them in profit after 250 dicerolls. Brown, train stations and the utilities generate poor profit and mightbe bad investments. But we noticed previously that yellow and red almosthad no increase in expected return after 3 houses and dark blue and greenactually had an decrease, hence it is interesting to see what happens whenwe stop with 3 houses for them.

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Figure 5: How the expected profit evolves over opponents turns were yel-low,green, dark blue and red have 3 houses and the rest have 5.

Here we have that dark blue, red, pink and yellow are very similar in bothprice and how much money they are expected to generate over time. Fromfigure 5 we see that red, dark blue and yellow are probably to prefer overpink since they perform as well with 3 houses as pink does with 5, makingthem more flexible and better in the long run since they can be developedfurther. Orange is very interesting here. It clearly has the best performanceand yet cost about as much pink, dark blue, yellow and red and is strictlybetter than green with lower cost and greater expected profit.

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4 Extended model of Monopoly

Even though it was possible to obtain interesting result from previous chap-ter the model does not involve an important feature of the game, the money.Furthermore it does not include picking up the streets and develop them asthe turns go. In this chapter the goal will be to try to incorporate moneyand test previous results. However when including money the states spacebecome infinite since the game rules does not have an upper limit on howmuch money there can be in a game. Even if the money was limited to befinite there is still all possible money for all players in the game times 120times all possible owned streets and houses. This is the motivation to whythis part will be more of a simulation.

4.1 The structure of the simulation

The first to consider was how a player would be represented. The way this wasdone was to represent it as a 3-tuple as before but extend it with money andway to keep track of what properties and houses the player has. The represen-tation looked as follows: (doubles, pos, timeJail, cash, streets) where streetskeeps track of which streets the player have with what amount of houses.A game of two players could now be represented as (player1, player2, turn)where turn was simply which players turn it was. So for example if player1starts then its (player1, player2, player1). Then after player1 plays her turnit became (player1, player2, player2) and so on. From this it was possibleto simulate 2-player games where different players took different decisions.It should be noted that in the simulations the cards will not be put backrandomly anymore put instead in bottom of the pile as the rules state.

4.2 Simulations

The first simulation was to compare the colours against each other and seehow they performed. It was of interest to look how the colours fared againsteach other with 3 houses since they had the most increase in return. Thesimulation was done as follows. Every player started at GO with 30 000 krand moved according to the rules. A player did only develop its chosen colourand did only develop to max 3 houses on each street she owned. Houses could

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only be bought if 3 times the cost of it was less than the money the playerowned. The reason for this was that so the player would not buy her selfruined/close to ruined. It should however be noted that the 3 is arbitrary.Every colour was simulated against every other 100 000 games.

colour win tie loss win probbrown 782 239324 459894 0.001blue 9582 347619 342799 0.014pink 100290 366928 232782 0.143

orange 250880 287291 161829 0.358red 330622 193946 175432 0.472

yellow 359704 156950 183346 0.514green 446849 59978 193173 0.638

dark blue 408447 133652 157901 0.583

Table 6: Colours with 3 houses simulated against each other.

Some key features to be taken away from the simulation is that firstly,brown, blue and pink performed poorly especially brown. This results agreewith the previous chapter. Moreover there seems to be a trend that themore expensive to better. However looking at orange losses it is actually thesecond lowest after dark blue. This indicates that orange might be a safe bet.

In similar manners a simulation was done for every pair of colours. Everypair met 15 different setups and each setup met each other 100 000 times.

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colour win ratedark blue and orange 67 %

orange and red 63 %red and dark blue 62 %orange and yellow 61 %pink and dark blue 60 %

yellow and dark blue/orange and green/ pink and orange 59 %green and dark blue 56 %

red and yellow/pink and red 55 %

Table 7: The top 11 performers.

In table 7 we can see that orange is in the best performing pair togetherwith dark blue and that orange is in 5 of 11 of the top performing pairs.Moreover there seems to be a trend that middle priced colours together withmore expensive colours work well. Lastly blue and brown does not appear inthe top 11.

5 Summary and discussion

To recap, let us recall the problem formulation of this paper. The mainquestion that was asked was: Does an optimal strategy of Monopolyexist? However since this was a rather difficult question to answer a heuristicwere made containing the following sub-questions :

• Do the probabilities of the tiles differ and which colour is most probableto land on?

• What is the optimal amount of houses to own for each street?

• Which colour generate most money?

• Which is the best colours to obtain?

There were differences in the probabilities of the tiles and the most prob-able colour to land on was orange. For every colour except dark blue andgreen the expected return of the houses increased for every bought houseup to 5 houses but the greatest increase in expected return was when going

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from 2 to 3 houses for every colour. When it comes to which colour gener-ates the most money it depends. Orange is the colour that is the quickest tomake profit but eventually green will generate most. However when having5 houses orange cost a lot less and will generate almost as much money asgreen. There is a even smaller difference between green, yellow and red butboth yellow and red are cheaper, but not as cheap as orange. Moreover or-ange seems to be the best colour. It is relative cheap, is the quickest to gainprofit and has one of the best the expected profit. From the simulations ithad the second least losses against all other colours and when looking at pairsit was in 5 out of the 11 best performing pairs including the best performingpair, dark blue and orange. Otherwise yellow, dark blue and red seems goodto buy. Green is probably too expensive and have to poor expected return toprioritize. Moreover blue seems to be reasonable to buy early since its cheapand have decent return.

It is still hard to give an answer if it exist an optimal strategy but someconclusions can be made. There are interesting differences in the coloursthat a player probably can use to gain advantage. For example based onthe simulation of pairs of colours a middle priced colour like orange or pinkseems to perform well combined with a more expensive street like dark blue.If looking at the gathered evidence one should probably skip brown, trainstations and utilities or have them as last resort. Since red and yellow barleyincreased in expected return and green and dark blue decreased in expectedreturn when increasing houses from 3 a player should probably stop invest-ing in them when she reach 3 houses and instead focus on developing othercolours further. Since these had very similar traits with pink with 5 housesthey are probably better to buy than pink since they can be develop furthermaking them more dynamic. However to be sure one would have to do amore in-depth analysis and the question still remain when to buy and inwhich order. More simulations should perhaps have been done that allowed5 houses and also test letting red, yellow, green and dark blue remain with 3houses until the other colour of the pair were fully developed. But then thequestion arise of how long do you want to wait and if you have money andare not in risk of becoming ruined then its probably best to invest the moneyanyway. Also it should be mentioned that just buying 1 or 2 colours arenot realistic in a 2-player game but the simulations nevertheless gave someintuition of which colours seems to be good.

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6 Further exploration

Although some results were made there is still features to explore towardsfinding good strategies. I think that a first thing to do would be to test moreof the results and trying out decision-policies based on what was discoveredin chapter 3. For example testing a strategy where if the player buys green,yellow, red or dark blue it only develop to 3 houses until finished developedother streets. Then test it without brown, utilities and train station andmaybe even exclude green due to it being very expensive. Then start tolook into blocks. With blocks I mean that even if a player are not interestedin a colour the player can block it from an opponent by buying one of thestreets so the opponent cannot develop houses there. However I think thesedecision rules might be to simple and miss a lot of important features like howmuch money does your opponent has, where is your opponent currently onthe board, what is her expected return, how does your decision change yourprobability of getting ruined and so on. In order to find more advance andbetter strategies I think Markov decision processes (MDP) are an interestingnext step. Intuitively a Markov decision process is a Markov process were thetransitions happens partly due to decisions, partly to chance. If possible tomodel the game as a MDP there exist ways to find the optimal decision-policy,namely the optimal strategy for the game. Another interesting extension is toincrease the amount of players. Monopoly is not a game that is often playedwith only 2 players but instead with maybe 4 or 5. By increasing the amountof players one could from there make some further natural extensions byallowing mortgaging properties, selling houses, trades, auctions and decisionsof how to get out of jail. These would make the game have more decisionswhich I think will increase the effect of making good decisions and makingthe game less about chance. For example it is easier to get the good streetsif they get auctioned out. From that you could for example also investigatehow much a player should be willing to pay for the street given that theplayer is in a certain state.

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References

[1] Richard F. Bass. Stochastic processes, volume 33;33.;. Cambridge Uni-versity Press, Cambridge;New York;, 2011.

[2] Gregory F. Lawler. Introduction to stochastic processes. Chapman Hal-l/CRC, Boca Raton, Fla, 2nd edition, 2006.

[3] David Stirzaker. Stochastic processes and models. Oxford Univ. Press,Oxford, 2005.